codeforce 985C

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C. Liebig's Barrels

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have m = n·k wooden staves. The i-th stave has length ai. You have to assemble n barrels consisting of k staves each, you can use any kstaves to construct a barrel. Each stave must belong to exactly one barrel.

Let volume vj of barrel j be equal to the length of the minimal stave in it.

You want to assemble exactly n barrels with the maximal total sum of volumes. But you have to make them equal enough, so a difference between volumes of any pair of the resulting barrels must not exceed l, i.e. |vx - vy| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.

Print maximal total sum of volumes of equal enough barrels or 0 if it's impossible to satisfy the condition above.

Input

The first line contains three space-separated integers n, k and l (1 ≤ n, k ≤ 105, 1 ≤ n·k ≤ 105, 0 ≤ l ≤ 109).

The second line contains m = n·k space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — lengths of staves.

Output

Print single integer — maximal total sum of the volumes of barrels or 0 if it's impossible to construct exactly n barrels satisfying the condition |vx - vy| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.

Examples

input

Copy

4 2 1
2 2 1 2 3 2 2 3

output

Copy

7

input

Copy

2 1 0
10 10

output

Copy

20

input

Copy

1 2 1
5 2

output

Copy

2

input

Copy

3 2 1
1 2 3 4 5 6

output

Copy

0

Note

In the first example you can form the following barrels: [1, 2], [2, 2], [2, 3], [2, 3].

In the second example you can form the following barrels: [10], [10].

In the third example you can form the following barrels: [2, 5].

In the fourth example difference between volumes of barrels in any partition is at least 2 so it is impossible to make barrels equal enough.

贪心（找到upper_bound(a,a+n,a[1])-1即为最后一个符合a[i]-a[1]的位置，对其前面查找符合条件的即可）

上代码<^..^>

#include 
#include
#include
#include
#include
using namespace std;
int n,k,l,i,cnt;
long long int sum;
int mark[100005],a[100005];
int main() {
	  memset(mark,0,sizeof(mark));
	  scanf("%d%d%d",&n,&k,&l);
	  for(i=1;i<=n*k;i++) scanf("%d",&a[i]);
	  sort(a+1,a+n*k+1);
	  int pos=upper_bound(a+1,a+n*k+1,a[1]+l)-(a+1); 
	  if(pos=1;i--)
	  	{
	  		if(!mark[i]) 
			  {
			  sum+=a[i];
			  cnt++;
	          }
	  		if(cnt==n) break;
		}
	  }
	  printf("%lld\n",sum);
	 return 0;
}

这道题有一个要注意的地方....upper_bound是下界，lower_bound是上界。。。。。。